Linear First Order ODE/(2 y - x^3) dx = x dy

Theorem

$(1): \quad \paren {2 y - x^3} \rd x = x \rd y$

$y = -x^3 + C x^2$

Rearranging $(1)$:

$(2): \quad \dfrac {\d y} {\d x} - \dfrac 2 x y = - x^2$

$(2)$ is in the form:

$\dfrac {\d y} {\d x} + \map P x y = \map Q x$

where $\map P x = -\dfrac 2 x$.

Thus:

$\ds \int \map P x \rd x$

$=$

$\ds \int -\dfrac 2 x \rd x$

$\ds $

$=$

$\ds -2 \ln x$

$\ds $

$=$

$\ds \map \ln {\frac 1 {x^2} }$

$\ds \leadsto \ \ $

$\ds e^{\int P \rd x}$

$=$

$\ds \frac 1 {x^2}$

Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:

$\map {\dfrac \d {\d x} } {\dfrac y {x^2} } = -1$

$\dfrac y {x^2} = -x + C$

or:

$y = -x^3 + C x^2$

$\blacksquare$